then I must prove that $f(e^{-\pi})=\frac1{24}$. It was not hard to find the relation between $f(x)$ and $g(x)$, namely $f(x)=g(x)-4g(x^2)+4g(x^4)$.

Note that $g(x)$ is a Lambert series, so by expanding the Taylor series for the denominators and reversing the two sums, I get

$$g(x)=\sum_{n=1}^{\infty}\sigma(n)x^n$$

where $\sigma$ is the divisor function $\sigma(n)=\sum_{d\mid n}d$.

I then define for complex $\tau$ the function
$$G_2(\tau)=\frac{\pi^2}3\Bigl(1-24\sum_{n=1}^{\infty}\sigma(n)e^{2\pi in\tau}\Bigr)$$ so that
$$f(e^{-\pi})=g(e^{-\pi})-4g(e^{-2\pi})+4g(e^{-4\pi})=\frac1{24}+\frac{-G_2(\frac i2)+4G_2(i)-4G_2(2i)}{8\pi^2}.$$

But it is proven in Apostol "Modular forms and Dirichlet Series", page 69-71 that $G_2\bigl(-\frac1{\tau}\bigr)=\tau^2G_2(\tau)-2\pi i\tau$, which gives $\begin{cases}G_2(i)=-G_2(i)+2\pi\\ G_2(\frac i2)=-4G_2(2i)+4\pi\end{cases}\quad$. This is exactly was needed to get the desired result.

Hitoshigoto oshimai !

I find that sum fascinating. $e,\pi$ all together to finally get a rational. This is why mathematics is beautiful!

I have seen you use similar techniques many times before. Where did you learn them? You are quite good with integrals.
–
PotatoMay 12 '13 at 7:32

1

@Potato: These techniques for summing infinite series are known in the literature as applications of integral transforms such as Mellin and Fourier transform. I have been learning them over the years. They are very effective techniques. See here for a Fourier transform technique for summing a series.
–
Mhenni BenghorbalMay 12 '13 at 7:48

Is it just me or is there an error in the signs of $$\sum_{k\ge 0} \frac{1}{(2k+1)^s} = \zeta(s) (1 - 2^{-s})$$ This error is repeated twice and canceled by the square, so that you still get the right answer.
–
Marko RiedelMay 13 '13 at 0:06

Actually the above is not quite complete, the missing piece is the proof that we can drop the contribution from the pole at $s=-1,$ which is $x/24.$ To verify this we have to show that
$$\int_{-i\infty}^{i\infty}
\frac{1}{\pi^{s+1}} \Gamma(s+1) (1-2^{-s})^2 \zeta(s+1)\zeta(s) ds = 0.$$
Now from the functional equation of the Riemann Zeta function we see that this integral is equal to
$$-\int_{-i\infty}^{i\infty}
\frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s})
\zeta(s) ds$$
Actually doing the accounting we find that the kernel
$$ g(s) = \frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s})
\zeta(s) $$ of this integral has the property that $g(s) = - g(-s)$ on the imaginary axis, so the integral is zero.

To see this consider what effect negation has on the individual terms.
$$\zeta(-s)\zeta(s) \to \zeta(s)\zeta(-s),$$
$$(2^s-1)(1-2^{-s}) \to (2^{-s}-1)(1-2^s) = (2^s-1)(1-2^{-s}),$$
$$\sin(1/2 s\pi) \to \sin(1/2 (-s)\pi) = -\sin(1/2 s\pi).$$
The first two terms are even and the last one is odd, QED.

Note that we have taken advantage of the fact that $x=1$ ... for other values of $x$ this trick will not go through. Also relevant is that negation (rotation by 180 degrees about the origin) takes the imaginary axis to itself (this is not the case when we are integrating along some other line parallel to the imaginary axis in the right half plane).

Let me rephrase my questions. 1) Can the right side of the rectangle be any vertical line to the right of the line $\Re(s) =1$? 2) There is a simple pole at the origin (albeit with residue 0). Does the contour technically need to be indented? 3) Does the integral go to zero along the top and bottom of the rectangle since $|\Gamma(s)|$ decays quickly as $\Im(s)$ increases?
–
Random VariableMay 16 '13 at 0:10

1

1) Owing to the convergence of $\sum_{k\ge 0} \frac{1}{(2k+1)^s}$ in the half plane $\Re(s)>1$ and the fact that there are no additional poles the Mellin inversion integral can indeed be along any vertical line in that half plane. 2) The simple pole is cancelled by the $(1-2^{-s})$ term, no indentation necessary. 3) This is correct, the decrease is exponential. Finally, let me refer you to one of the experts on this one -- the paper "Mellin Transform and Its Applications" by Szpankowski on academia.edu contains many examples and is highly readable. (For some reason SE won't let me add a link.)
–
Marko RiedelMay 16 '13 at 1:21